given the verticeS P(-5,2), Q(-1,-5), R(9,10) prove uSing Slopes that PQR is a right triangle
Find triangle's sides and they should be: a<b+c b<a+c c<a+b
find the slopes: 1) between P and Q 2) between Q and R 3) between P and R if two lines are perpendicular to each other, then the product of their slopes = -1 use this fact to show PQR is a right triangle.
there are many ways to skin a cat
is slope is same word as side?
no
slope = rise/run
Things get tricky when you find that one of the slopes is NaN
|dw:1322315207346:dw| so PR PQ RQ is sides and slopes isn't it?
well, each side has a slope associated with it :-P
for two point with coordinate (x1, y1) and (x2, y2), slope = (y2-y1)/(x2-x1) side = distance between these two points
well what do you get ?
\(y=\mbox{k}x+\mbox{b}\) k is slope right?
yes
@mikow - the 3 slopes I get are \(\frac{-7}{4}, \frac{3}{2} and \frac{4}{7}\)
the products of the slopes -7/4 and 4/7 is -1, so the two lines with those slopes denote the right angle of the right triangle
@asnaseer .. we're the Same :)
what about poor @agdgdgdgwngo :(
good - glad you got there in the end :-) @agdg... (why is your name tag so difficult to type) - yes you were also right :-)
dont be Sad :)
be happy like JOLLIBEE AHAHAHHA LOL
it;s easy to type:: 1. place your left pinky on the shift button, your left ring-finger on the a button, and your index finger on the g button. 2. press shift and a 3. release shift 4. press g 5. move your middle finger over the d button. 6 repeat three times: press the d button and the g button, in order 7. type in "wngo"
LOL! - do you hold special training sessions that I can attend? ;-)
yeah only 50 (insert your currency) per hour!
tnx to both of you :)
@agdgdgdgwngo - JUST in case case your name really is agdgdgdgwngo - I mean no offence by my remarks - it's all in jest. and I can only type with my index fingers as I never mastered touch typing :-(
no shame in hunt-and-peck style typing : http://en.wikipedia.org/wiki/Typing#Hunt_and_peck
lol^
Wow! I never new there was a term for my disability ;-)
wait, its a disablility
:-D
lets just call it a "special" talent shall we :-)
|dw:1322317814903:dw| show that the nearest distance from the common point to the plane is lA.BxCl / lAxB+BxC+CxAl
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